Given a complete binary tree, count the number of nodes.
Note:
Definition of a complete binary tree from :
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.
Example:
Input:
1
/ \
2 3
/ \ /
4 5 6
Output: 6
Solution
1. level order traversal
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int countNodes(TreeNode* root) {
if (root == nullptr) return 0;
int count = 0;
vector<TreeNode*> current({root});
vector<TreeNode*> next;
while (!current.empty()) {
next.reserve(current.size() * 2);
count += current.size();
for (const auto n: current) {
if (n->left != nullptr) next.push_back(n->left);
else break;
if (n->right != nullptr) next.push_back(n->right);
else break;
}
current.clear();
current.swap(next);
}
return count;
}
};
2. Recursive
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int countNodes(TreeNode* root) {
if (root == nullptr) return 0;
const int hl = height(root->left), hr = height(root->right);
if (hl == hr) return (1 << hl) + countNodes(root->right);
return (1 << hr) + countNodes(root->left);
}
inline int height(TreeNode *root) {
if (root == nullptr) return 0;
int h = 0;
while (root != nullptr) {
++h;
root = root->left;
}
return h;
}
};
